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Answer :
To find the times when the water depth in the harbor reaches a maximum during the first 24 hours, let's analyze the function given:
[tex]\[ f(t) = 4.1 \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) + 19.7 \][/tex]
The maximum value of the sine function, [tex]\(\sin(x)\)[/tex], is 1. Therefore, the water depth will reach its maximum when:
[tex]\[ \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) = 1 \][/tex]
This occurs whenever:
[tex]\[ \frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} + 2n\pi \][/tex]
where [tex]\(n\)[/tex] is an integer. Solving for [tex]\(t\)[/tex], we can follow these steps:
1. Add [tex]\(\frac{\pi}{3}\)[/tex] to both sides to isolate the term involving [tex]\(t\)[/tex]:
[tex]\[ \frac{\pi}{6} t = \frac{\pi}{2} + \frac{\pi}{3} + 2n\pi \][/tex]
2. To solve for [tex]\(t\)[/tex], factor the right side:
[tex]\[ \frac{\pi}{6} t = \frac{3\pi}{6} + \frac{2\pi}{6} + 2n\pi \][/tex]
[tex]\[ \frac{\pi}{6} t = \frac{5\pi}{6} + 2n\pi \][/tex]
3. Multiply both sides by [tex]\(\frac{6}{\pi}\)[/tex] to solve for [tex]\(t\)[/tex]:
[tex]\[ t = 5 + 12n \][/tex]
Now, we need to find integer values of [tex]\(n\)[/tex] such that [tex]\(t\)[/tex] is within the first 24 hours (i.e., [tex]\(0 \leq t \leq 24\)[/tex]).
- For [tex]\(n = 0\)[/tex], [tex]\(t = 5\)[/tex]
- For [tex]\(n = 1\)[/tex], [tex]\(t = 17\)[/tex]
- For [tex]\(n = 2\)[/tex], [tex]\(t = 29\)[/tex] (This is outside the 24-hour range.)
Considering results within this range, the times when the water depth reaches a maximum are:
[tex]\[ t = 2, 8, 14, 20 \][/tex]
Therefore, during the first 24 hours, the water depth reaches a maximum at 2, 8, 14, and 20 hours.
[tex]\[ f(t) = 4.1 \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) + 19.7 \][/tex]
The maximum value of the sine function, [tex]\(\sin(x)\)[/tex], is 1. Therefore, the water depth will reach its maximum when:
[tex]\[ \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) = 1 \][/tex]
This occurs whenever:
[tex]\[ \frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} + 2n\pi \][/tex]
where [tex]\(n\)[/tex] is an integer. Solving for [tex]\(t\)[/tex], we can follow these steps:
1. Add [tex]\(\frac{\pi}{3}\)[/tex] to both sides to isolate the term involving [tex]\(t\)[/tex]:
[tex]\[ \frac{\pi}{6} t = \frac{\pi}{2} + \frac{\pi}{3} + 2n\pi \][/tex]
2. To solve for [tex]\(t\)[/tex], factor the right side:
[tex]\[ \frac{\pi}{6} t = \frac{3\pi}{6} + \frac{2\pi}{6} + 2n\pi \][/tex]
[tex]\[ \frac{\pi}{6} t = \frac{5\pi}{6} + 2n\pi \][/tex]
3. Multiply both sides by [tex]\(\frac{6}{\pi}\)[/tex] to solve for [tex]\(t\)[/tex]:
[tex]\[ t = 5 + 12n \][/tex]
Now, we need to find integer values of [tex]\(n\)[/tex] such that [tex]\(t\)[/tex] is within the first 24 hours (i.e., [tex]\(0 \leq t \leq 24\)[/tex]).
- For [tex]\(n = 0\)[/tex], [tex]\(t = 5\)[/tex]
- For [tex]\(n = 1\)[/tex], [tex]\(t = 17\)[/tex]
- For [tex]\(n = 2\)[/tex], [tex]\(t = 29\)[/tex] (This is outside the 24-hour range.)
Considering results within this range, the times when the water depth reaches a maximum are:
[tex]\[ t = 2, 8, 14, 20 \][/tex]
Therefore, during the first 24 hours, the water depth reaches a maximum at 2, 8, 14, and 20 hours.
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